Find the largest real number $c$ such that \[x_1^2 + x_2^2 + \dots + x_{101}^2 \geq cM^2\]whenever $x_1,x_2,\ldots,x_{101}$ are real numbers such that $x_1+x_2+\cdots+x_{101}=0$ and $M$ is the median of $x_1,x_2,\ldots,x_{101}.$
Solution: Since the inequality is always true for $M = 0,$ it suffices to consider the case $M \neq 0.$

For a particular $c$ and for any tuple $(x_1, \dots, x_{101})$ satisfying the conditions, the tuple $(-x_1, \dots, -x_{101})$ satisfies the conditions as well, so we may assume that $M > 0.$ Finally, we may assume that $x_1 \le x_2 \le \dots \le x_{101},$ so that $M = x_{51}.$

We want to find the largest $c$ such that the inequality \[x_1^2 + x_2^2 + \dots + x_{101}^2 \ge cx_{51}^2\]always holds, where $x_1 \le x_2 \le \dots \le x_{101}$ and $x_1 + x_2 + \dots + x_{101} = 0.$ Therefore, fixing a value of $x_{51},$ we should write inequalities that minimize $x_1^2 + x_2^2 + \dots + x_{101}^2.$

To compare the terms on the left-hand side to $x_{51}^2,$ we deal with the terms $x_1^2 + x_2^2 + \dots + x_{50}^2$ and $x_{51}^2+x_{52}^2+\dots+x_{101}^2$ separately.

By Cauchy-Schwarz, \[(1 + 1 + \dots + 1)(x_1^2+x_2^2+\dots+x_{50}^2) \ge (x_1+x_2+\dots+x_{50})^2,\]so \[x_1^2 + x_2^2 + \dots + x_{50}^2 \ge \tfrac{1}{50}\left(x_1+x_2+\dots+x_{50}\right)^2.\]We have $x_1+x_2+\dots+x_{50} = -x_{51}-x_{52} -\dots - x_{101}\le -51x_{51} $ because $x_{51} \le x_{52} \le \dots \le x_{101}.$ Since $x_{51} > 0,$ both $x_1 + x_2 + \dots + x_{50}$ and $-51x_{51}$ are negative, so we can write \[\begin{aligned} x_1^2+x_2^2+\dots+x_{50}^2 &\ge \tfrac{1}{50} (x_1+x_2+\dots+x_{50})^2\\ & \ge\tfrac{1}{50} \left(-51x_{51}\right)^2 \\ &= \tfrac{51^2}{50} x_{51}^2. \end{aligned}\]On the other hand, since $0 < x_{51} \le x_{52} \le \dots \le x_{101},$ we simply have \[x_{51}^2 + x_{52}^2 + \dots + x_{101}^2 \ge 51x_{51}^2.\]Putting all this together gives \[(x_1^2 + x_2^2 + \dots + x_{50})^2 + (x_{51}^2 + x_{52}^2 + \dots + x_{101}^2) \ge \left(\tfrac{51^2}{50} + 51\right) x_{51}^2 = \tfrac{5151}{50} x_{51}^2.\]Equality holds when $x_1 = x_2 = \dots = x_{50} = -\tfrac{51}{50}$ and $x_{51} = x_{52} = \dots = x_{101} = 1,$ so the answer is $\boxed{\tfrac{5151}{50}}.$